Question

three identical point masses; each of mass 1kg lie in the x - y plane at points (0,0),(0,0.2m) and (0.2m,0). the net gravitational force on the mass at the origin is 1) 1.67×10^(-9) (\hat{i}+\hat{j})n

Explanation:

Step1: Recall gravitational - force formula

The gravitational force between two point - masses $m_1$ and $m_2$ separated by a distance $r$ is given by $\vec{F}=G\frac{m_1m_2}{r^2}\hat{r}$, where $G = 6.67\times10^{-11}\ Nm^2/kg^2$.

Step2: Calculate force due to mass at $(0,0.2m)$

Let $m_1 = m_2=1\ kg$ and $r = 0.2m$. The force $\vec{F}_1$ on the mass at the origin $(0,0)$ due to the mass at $(0,0.2m)$ is $\vec{F}_1=G\frac{m\times m}{(0.2)^2}\hat{j}$, where $G = 6.67\times10^{-11}\ Nm^2/kg^2$ and $m = 1\ kg$. So, $\vec{F}_1=6.67\times10^{-11}\times\frac{1\times1}{(0.2)^2}\hat{j}= 1.6675\times10^{-9}\hat{j}\ N$.

Step3: Calculate force due to mass at $(0.2m,0)$

The force $\vec{F}_2$ on the mass at the origin $(0,0)$ due to the mass at $(0.2m,0)$ is $\vec{F}_2=G\frac{m\times m}{(0.2)^2}\hat{i}$. Substituting $G = 6.67\times10^{-11}\ Nm^2/kg^2$ and $m = 1\ kg$, we get $\vec{F}_2=6.67\times10^{-11}\times\frac{1\times1}{(0.2)^2}\hat{i}=1.6675\times10^{-9}\hat{i}\ N$.

Step4: Calculate net force

The net force $\vec{F}=\vec{F}_1+\vec{F}_2$. So, $\vec{F}=1.67\times10^{-9}(\hat{i}+\hat{j})\ N$.

Answer:

$1.67\times10^{-9}(\hat{i}+\hat{j})\ N$